Given N jobs where every job is represented by following three elements of it.
- Start Time
- Finish Time
- Profit or Value Associated
Find the maximum profit subset of jobs such that no two jobs in the subset overlap.
Example:
Input: Number of Jobs n = 4
Job Details {Start Time, Finish Time, Profit}
Job 1: {1, 2, 50}
Job 2: {3, 5, 20}
Job 3: {6, 19, 100}
Job 4: {2, 100, 200}
Output: The maximum profit is 250.
We can get the maximum profit by scheduling jobs 1 and 4.
Note that there is longer schedules possible Jobs 1, 2 and 3
but the profit with this schedule is 20+50+100 which is less than 250.
We strongly recommend to refer below article as a prerequisite for this.
Weighted Job Scheduling
The above problem can be solved using following recursive solution.
1) First sort jobs according to finish time.
2) Now apply following recursive process.
// Here arr[] is array of n jobs
findMaximumProfit(arr[], n)
{
a) if (n == 1) return arr[0];
b) Return the maximum of following two profits.
(i) Maximum profit by excluding current job, i.e.,
findMaximumProfit(arr, n-1)
(ii) Maximum profit by including the current job
}
How to find the profit including current job?
The idea is to find the latest job before the current job (in
sorted array) that doesn't conflict with current job 'arr[n-1]'.
Once we find such a job, we recur for all jobs till that job and
add profit of current job to result.
In the above example, "job 1" is the latest non-conflicting
for "job 4" and "job 2" is the latest non-conflicting for "job 3".
We have discussed recursive and Dynamic Programming based approaches in the previous article. The implementations discussed in above post uses linear search to find the previous non-conflicting job. In this post, Binary Search based solution is discussed. The time complexity of Binary Search based solution is O(n Log n).
The algorithm is:
- Sort the jobs by non-decreasing finish times.
- For each i from 1 to n, determine the maximum value of the schedule from the subsequence of jobs[0..i]. Do this by comparing the inclusion of job[i] to the schedule to the exclusion of job[i] to the schedule, and then taking the max.
To find the profit with inclusion of job[i]. we need to find the latest job that doesn’t conflict with job[i]. The idea is to use Binary Search to find the latest non-conflicting job.
C++
// C++ program for weighted job scheduling using Dynamic // Programming and Binary Search #include <iostream> #include <algorithm> using namespace std; // A job has start time, finish time and profit. struct Job { int start, finish, profit; }; // A utility function that is used for sorting events // according to finish time bool myfunction(Job s1, Job s2) { return (s1.finish < s2.finish); } // A Binary Search based function to find the latest job // (before current job) that doesn't conflict with current // job. "index" is index of the current job. This function // returns -1 if all jobs before index conflict with it. // The array jobs[] is sorted in increasing order of finish // time. int binarySearch(Job jobs[], int index) { // Initialize 'lo' and 'hi' for Binary Search int lo = 0, hi = index - 1; // Perform binary Search iteratively while (lo <= hi) { int mid = (lo + hi) / 2; if (jobs[mid].finish <= jobs[index].start) { if (jobs[mid + 1].finish <= jobs[index].start) lo = mid + 1; else return mid; } else hi = mid - 1; } return -1; } // The main function that returns the maximum possible // profit from given array of jobs int findMaxProfit(Job arr[], int n) { // Sort jobs according to finish time sort(arr, arr+n, myfunction); // Create an array to store solutions of subproblems. table[i] // stores the profit for jobs till arr[i] (including arr[i]) int *table = new int[n]; table[0] = arr[0].profit; // Fill entries in table[] using recursive property for (int i=1; i<n; i++) { // Find profit including the current job int inclProf = arr[i].profit; int l = binarySearch(arr, i); if (l != -1) inclProf += table[l]; // Store maximum of including and excluding table[i] = max(inclProf, table[i-1]); } // Store result and free dynamic memory allocated for table[] int result = table[n-1]; delete[] table; return result; } // Driver program int main() { Job arr[] = {{3, 10, 20}, {1, 2, 50}, {6, 19, 100}, {2, 100, 200}}; int n = sizeof(arr)/sizeof(arr[0]); cout << "Optimal profit is " << findMaxProfit(arr, n); return 0; } |
Java
// Java program for Weighted Job Scheduling in O(nLogn)// timeimport java.util.Arrays;import java.util.Comparator;// Class to represent a jobclass Job{ int start, finish, profit; // Constructor Job(int start, int finish, int profit) { this.start = start; this.finish = finish; this.profit = profit; }}// Used to sort job according to finish timesclass JobComparator implements Comparator<Job>{ public int compare(Job a, Job b) { return a.finish < b.finish ? -1 : a.finish == b.finish ? 0 : 1; }}public class WeightedIntervalScheduling{ /* A Binary Search based function to find the latest job (before current job) that doesn't conflict with current job. "index" is index of the current job. This function returns -1 if all jobs before index conflict with it. The array jobs[] is sorted in increasing order of finish time. */ static public int binarySearch(Job jobs[], int index) { // Initialize 'lo' and 'hi' for Binary Search int lo = 0, hi = index - 1; // Perform binary Search iteratively while (lo <= hi) { int mid = (lo + hi) / 2; if (jobs[mid].finish <= jobs[index].start) { if (jobs[mid + 1].finish <= jobs[index].start) lo = mid + 1; else return mid; } else hi = mid - 1; } return -1; } // The main function that returns the maximum possible // profit from given array of jobs static public int schedule(Job jobs[]) { // Sort jobs according to finish time Arrays.sort(jobs, new JobComparator()); // Create an array to store solutions of subproblems. // table[i] stores the profit for jobs till jobs[i] // (including jobs[i]) int n = jobs.length; int table[] = new int[n]; table[0] = jobs[0].profit; // Fill entries in M[] using recursive property for (int i=1; i<n; i++) { // Find profit including the current job int inclProf = jobs[i].profit; int l = binarySearch(jobs, i); if (l != -1) inclProf += table[l]; // Store maximum of including and excluding table[i] = Math.max(inclProf, table[i-1]); } return table[n-1]; } // Driver method to test above public static void main(String[] args) { Job jobs[] = {new Job(1, 2, 50), new Job(3, 5, 20), new Job(6, 19, 100), new Job(2, 100, 200)}; System.out.println("Optimal profit is " + schedule(jobs)); }} |
Python
# Python program for weighted job scheduling using Dynamic # Programming and Binary Search# Class to represent a jobclass Job: def __init__(self, start, finish, profit): self.start = start self.finish = finish self.profit = profit# A Binary Search based function to find the latest job# (before current job) that doesn't conflict with current# job. "index" is index of the current job. This function# returns -1 if all jobs before index conflict with it.# The array jobs[] is sorted in increasing order of finish# time.def binarySearch(job, start_index): # Initialize 'lo' and 'hi' for Binary Search lo = 0 hi = start_index - 1 # Perform binary Search iteratively while lo <= hi: mid = (lo + hi) // 2 if job[mid].finish <= job[start_index].start: if job[mid + 1].finish <= job[start_index].start: lo = mid + 1 else: return mid else: hi = mid - 1 return -1# The main function that returns the maximum possible# profit from given array of jobsdef schedule(job): # Sort jobs according to finish time job = sorted(job, key = lambda j: j.finish) # Create an array to store solutions of subproblems. table[i] # stores the profit for jobs till arr[i] (including arr[i]) n = len(job) table = [0 for _ in range(n)] table[0] = job[0].profit; # Fill entries in table[] using recursive property for i in range(1, n): # Find profit including the current job inclProf = job[i].profit l = binarySearch(job, i) if (l != -1): inclProf += table[l]; # Store maximum of including and excluding table[i] = max(inclProf, table[i - 1]) return table[n-1]# Driver code to test above functionjob = [Job(1, 2, 50), Job(3, 5, 20), Job(6, 19, 100), Job(2, 100, 200)]print("Optimal profit is"),print schedule(job) |
C#
using System;using System.Collections.Generic;class Job { public int start, finish, profit; // Constructor public Job(int start, int finish, int profit) { this.start = start; this.finish = finish; this.profit = profit; }}// Used to sort job according to finish timesclass JobComparer : IComparer<Job> { public int Compare(Job a, Job b) { return a.finish < b.finish ? -1 : a.finish == b.finish ? 0 : 1; }}class WeightedIntervalScheduling { /* A Binary Search based function to find the latest job (before current job) that doesn't conflict with current job. "index" is index of the current job. This function returns -1 if all jobs before index conflict with it. The array jobs[] is sorted in increasing order of finish time. */ static public int BinarySearch(List<Job> jobs, int index) { // Initialize 'lo' and 'hi' for Binary Search int lo = 0, hi = index - 1; // Perform binary Search iteratively while (lo <= hi) { int mid = (lo + hi) / 2; if (jobs[mid].finish <= jobs[index].start) { if (jobs[mid + 1].finish <= jobs[index].start) lo = mid + 1; else return mid; } else hi = mid - 1; } return -1; } // The main function that returns the maximum possible // profit from given array of jobs static public int Schedule(List<Job> jobs) { // Sort jobs according to finish time jobs.Sort(new JobComparer()); // Create an array to store solutions of // subproblems. table[i] stores the profit for jobs // till jobs[i] (including jobs[i]) int n = jobs.Count; int[] table = new int[n]; table[0] = jobs[0].profit; // Fill entries in M[] using recursive property for (int i = 1; i < n; i++) { // Find profit including the current job int inclProf = jobs[i].profit; int l = BinarySearch(jobs, i); if (l != -1) inclProf += table[l]; // Store maximum of including and excluding table[i] = Math.Max(inclProf, table[i - 1]); } return table[n - 1]; } // Driver method to test above public static void Main(string[] args) { List<Job> jobs = new List<Job>{ new Job(1, 2, 50), new Job(3, 5, 20), new Job(6, 19, 100), new Job(2, 100, 200) }; Console.WriteLine("Optimal profit is " + Schedule(jobs)); }}// This code is contributed by ishankhandelwals. |
Javascript
<script>// JavaScript program for weighted job scheduling using Dynamic// Programming and Binary Search// Class to represent a jobclass Job{ constructor(start, finish, profit){ this.start = start this.finish = finish this.profit = profit }}// A Binary Search based function to find the latest job// (before current job) that doesn't conflict with current// job. "index" is index of the current job. This function// returns -1 if all jobs before index conflict with it.// The array jobs[] is sorted in increasing order of finish// time.function binarySearch(job, start_index){ // Initialize 'lo' and 'hi' for Binary Search let lo = 0 let hi = start_index - 1 // Perform binary Search iteratively while(lo <= hi){ let mid = Math.floor((lo + hi) /2); if (job[mid].finish <= job[start_index].start){ if(job[mid + 1].finish <= job[start_index].start) lo = mid + 1 else return mid } else hi = mid - 1 } return -1}// The main function that returns the maximum possible// profit from given array of jobsfunction schedule(job){ // Sort jobs according to finish time job.sort((a,b)=>a.finish - b.finish) // Create an array to store solutions of subproblems. table[i] // stores the profit for jobs till arr[i] (including arr[i]) let n = job.length let table = new Array(n).fill(0) table[0] = job[0].profit; // Fill entries in table[] using recursive property for(let i=1;i<n;i++){ // Find profit including the current job inclProf = job[i].profit l = binarySearch(job, i) if (l != -1) inclProf += table[l]; // Store maximum of including and excluding table[i] = Math.max(inclProf, table[i - 1]) } return table[n-1]}// Driver code to test above functionlet job = [new Job(1, 2, 50), new Job(3, 5, 20), new Job(6, 19, 100), new Job(2, 100, 200)] document.write("Optimal profit is ")document.write(schedule(job),"</br>")// This code is contributed by shinjanpatra.</script> |
Output:
Optimal profit is 250
Time complexity: O(n Log n)
Auxiliary Space: O(n) because using extra space for array table
This article is contributed by Daniel Ray. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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